Cho a,b,c khác 0 và 1/a+ 1/b+1/c = 0 CMR $\sqrt{a+b}=\sqrt{a+c}+\sqrt{b+c}$

Question

Cho a,b,c khác 0 và 1/a+ 1/b+1/c = 0
CMR $\sqrt{a+b}=\sqrt{a+c}+\sqrt{b+c}$

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Valerie 1 tháng 2021-08-11T04:23:58+00:00 2 Answers 5 views 0

Answers ( )

    0
    2021-08-11T04:25:02+00:00

    Giải thích các bước giải:

    Ta có:

    $\sqrt{a+b}=\sqrt{a+c}+\sqrt{b+c}$

    $\leftrightarrow \sqrt{a+b}-\sqrt{a+c}=\sqrt{b+c}$

    $\leftrightarrow (\sqrt{a+b}-\sqrt{a+c})^2=(\sqrt{b+c})^2$

    $\leftrightarrow a+b+a+c-2\cdot\sqrt{a+b}\cdot\sqrt{a+c}=b+c$

    $\leftrightarrow 2a=2\cdot\sqrt{(a+b)(a+c)}$

    $\leftrightarrow \sqrt{(a+b)(a+c)}=a$

    $\leftrightarrow (a+b)(a+c)=a^2$

    $\leftrightarrow a^2+ab+ac+bc=a^2$

    $\leftrightarrow ab+ac+bc=0$

    $\leftrightarrow \dfrac{ab+ac+bc}{abc}=0$

    $\leftrightarrow \dfrac1a+\dfrac1b+\dfrac1c=0$ (đúng)

    $\to$ đpcm

    0
    2021-08-11T04:25:47+00:00

    Xét $\sqrt[]{a+b} = \sqrt[]{a+c} + \sqrt[]{b+c}$

    $\to a+b= a+c+b+c+2\sqrt[]{(a+c).(b+c)}$

    $\to a+b=a+b+2c+2\sqrt[]{(a+c).(b+c)}$

    $\to -c = \sqrt[]{(a+c).(b+c)}$

    $\to (-c)^2 = (a+c).(b+c)$

    $\to c^2 = ab+ac+bc+c^2$

    $\to ab+bc+ca= 0 $

    $\to \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} = 0$ ( Đúng )

    $\to đpcm$

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